What is the derivative of 7^{-4x+1+2x^{-1}} ?

The derivative of 7^{-4x+1+2x^{-1}} is ln(7)*7^{-4x+1+2/x}(-2/(x^2)-4)

What is the first derivative of 7^{-4x+1+2x^{-1}} ?

The first derivative of 7^{-4x+1+2x^{-1}} is ln(7)*7^{-4x+1+2/x}(-2/(x^2)-4)

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\frac{d}{d x} (7^{- 4 x + 1 + 2 x^{- 1}})

ln (7) \cdot 7^{- 4 x + 1 + \frac{2}{x}} (- \frac{2}{x ^{2}} - 4)

Solution steps

\frac{d}{d x} (7^{- 4 x + 1 + 2 x^{- 1}})

Apply exponent rule:

a^{b} = e^{b l n (a)}

7^{- 4 x + 1 + 2 x^{- 1}} = e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)}

= \frac{d}{d x} (e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)})

Apply the chain rule:

e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)} \frac{d}{d x} ((- 4 x + 1 + 2 x^{- 1}) ln (7))

= e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)} \frac{d}{d x} ((- 4 x + 1 + 2 x^{- 1}) ln (7))

\frac{d}{d x} ((- 4 x + 1 + 2 x^{- 1}) ln (7)) = ln (7) (- \frac{2}{x ^{2}} - 4)

= e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)} ln (7) (- \frac{2}{x ^{2}} - 4)

e^{(- 4 x + 1 + 2 x^{- 1}) l n (7)} = 7^{- 4 x + 1 + \frac{2}{x}}

= ln (7) \cdot 7^{- 4 x + 1 + \frac{2}{x}} (- \frac{2}{x ^{2}} - 4)

\frac{d}{d x} (x (- \frac{1}{10} x + 610))

x \to 1 lim (2^{0})

\int_{2}^{3} 1 d x

\frac{x ^{2} - 9}{x - 3}

\frac{( s + 7 )}{s ^{2} + 11 s + 30}

What is the derivative of 7^{-4x+1+2x^{-1}} ?
The derivative of 7^{-4x+1+2x^{-1}} is ln(7)*7^{-4x+1+2/x}(-2/(x^2)-4)
What is the first derivative of 7^{-4x+1+2x^{-1}} ?
The first derivative of 7^{-4x+1+2x^{-1}} is ln(7)*7^{-4x+1+2/x}(-2/(x^2)-4)